Section 1.14 : Absolute Value Equations and Inequalities

Solve each of the following. Show All Solutions Hide All Solutions

1. \(\left| {3x + 8} \right| = 2\)
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   This uses the following fact

   \[\left| p \right| = d \ge 0\,\,\,\,\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,p = \pm \,d\]

   The requirement that \(d\) be greater than or equal to zero is simply an acknowledgement that absolute value only returns number that are greater than or equal to zero. See Problem 3 below to see what happens when \(d\) is negative.

   So, the solution to this equation is

   \[\begin{align*} \begin{aligned}3x + 8 & = 2\\ 3x & = - 6\\ x & = - 2\end{aligned} & {\hspace{0.75in}{\rm{OR}}\hspace{0.5in}} & \begin{aligned}3x + 8 & = - 2\\ 3x & = - 10\\ x & = - \frac{{10}}{3}\end{aligned}\end{align*}\]

   So, there were two solutions to this. That will almost always be the case. Also, do not get excited about the fact that these solutions are negative. This is not a problem. We can plug negative numbers into an absolute value equation (which is what we’re doing with these answers), we just can’t get negative numbers out of an absolute value (which we don’t, we get 2 out of the absolute value in this case).

2. \(\left| {2x - 4} \right| = 10\)
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   This one works identically to the previous problem.

   \[\begin{align*}\begin{aligned}2x - 4 & = 10\\ 2x & = 14\\ x & = 7\end{aligned} & \hspace{0.75in}{\rm{OR}}\hspace{0.5in} & \begin{aligned}2x - 4 & = - 10\\ 2x & = - 6\\ x & = - 3\end{aligned}\end{align*}\]

   Do not make the following very common mistake in solve absolute value equations and inequalities.

   \[\begin{align*}\left| {2x - 4} \right| \ne 2x + 4 & = 10\\ 2x & = 6\\ x & = 3\end{align*}\]

   Did you catch the mistake? In dropping the absolute value bars I just changed every “-” into a “+” and we know that doesn’t work that way! By doing this we get a single answer and it’s incorrect as well. Simply plug it into the original equation to convince yourself that it’s incorrect.

   \[\left| {2\left( 3 \right) - 4} \right| = \left| {6 - 4} \right| = \left| 2 \right| = 2 \ne 10\]

   When first learning to solve absolute value equations and inequalities people tend to just convert all minus signs to plus signs and solve. This is simply incorrect and will almost never get the correct answer. The way to solve absolute value equations is the way that I’ve shown here.

3. \(\left| {x + 1} \right| = - 15\)
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   This question is designed to make sure you understand absolute values. In this case we are after the values of \(x\) such that when we plug them into \(\left| {x + 1} \right|\) we will get -15. This is a problem however. Recall that absolute value ALWAYS returns a positive number! In other words, there is no way that we can get -15 out of this absolute value. Therefore, there are no solutions to this equation.

4. \(\left| {7x - 10} \right| \le 4\)
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   To solve absolute value inequalities with < or \( \le \) in them we use

   \[\begin{align*}\left| p \right| < d\,\,\,\,\hspace{0.25in}\,\,\,\, & \Rightarrow \hspace{0.25in}\,\,\,\,\,\, - d < p < d\\ \left| p \right| \le d\,\,\,\,\hspace{0.25in}\,\,\,\, & \Rightarrow \hspace{0.25in}\,\,\,\,\,\, - d \le p \le d\end{align*}\]

   As with absolute value equations we will require that d be a number that is greater than or equal to zero.

   The solution in this case is then

   \[\begin{array}{c} - 4 \le 7x - 10 \le 4\\ 6 \le 7x \le 14\\ \frac{6}{7} \le x \le 2\end{array}\]

   In solving these make sure that you remember to add the 10 to BOTH sides of the inequality and divide BOTH sides by the 7. One of the more common mistakes here is to just add or divide one side.

5. \(\left| {1 - 2x} \right| < 7\)
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   This one is identical to the previous problem with one small difference.

   \[\begin{array}{c} - 7 < 1 - 2x < 7\\ - 8 < - 2x < 6\\ 4 > x > - 3\end{array}\]

   Don’t forget that when multiplying or dividing an inequality by a negative number (-2 in this case) you’ve got to flip the direction of the inequality.

6. \(\left| {x - 9} \right| \le - 1\)
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   This problem is designed to show you how to deal with negative numbers on the other side of the inequality. So, we are looking for \(x\)’s which will give us a number (after taking the absolute value of course) that will be less than -1, but as with Problem 3 this just isn’t possible since absolute value will always return a positive number or zero neither of which will ever be less than a negative number. So, there are no solutions to this inequality.

7. \(\left| {4x + 5} \right| > 3\)
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   Absolute value inequalities involving > and \( \ge \) are solved as follows.

   \[\begin{align*}\left| p \right| > d\,\,\,\,\hspace{0.25in}\,\,\,\, & \Rightarrow \hspace{0.25in}\,\,\,\,\,\,p < - d\hspace{0.25in}{\rm{or}}\hspace{0.25in}p > d\\ \left| p \right| \ge d\,\,\,\,\hspace{0.25in}\,\,\,\, & \Rightarrow \hspace{0.25in}\,\,\,\,\,\,p \le - d\hspace{0.25in}{\rm{or}}\hspace{0.25in}p \ge d\end{align*}\]

   Note that you get two separate inequalities in the solution. That is the way that it must be. You can NOT put these together into a single inequality. Once I get the solution to this problem I’ll show you why that is.

   Here is the solution

   \[\begin{align*}\begin{aligned}4x + 5 & < - 3\\ 4x & < - 8\\ x & < - 2\end{aligned} & \hspace{0.75in}{\rm{OR}}\hspace{0.5in} & \begin{aligned}4x + 5 & > 3\\ 4x & > - 2\\ x & > - \frac{1}{2}\end{aligned}\end{align*}\]

   So the solution to this inequality will be \(x\)’s that are less than -2 or greater than \( - \frac{1}{2}\).

   Now, as I mentioned earlier you CAN NOT write the solution as the following double inequality.

   \[ - 2 > x > - \frac{1}{2}\]

   When you write a double inequality (as we have here) you are saying that \(x\) will be a number that will simultaneously satisfy both parts of the inequality. In other words, in writing this I’m saying that \(x\) is some number that is less than -2 and AT THE SAME TIME is greater than \( - \frac{1}{2}\). I know of no number for which this is true. So, this is simply incorrect. Don’t do it. This is however, a VERY common mistake that students make when solving this kind of inequality.

8. \(\left| {4 - 11x} \right| \ge 9\)
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   Not much to this solution. Just be careful when you divide by the -11.

   \[\begin{align*}\begin{aligned}4 - 11x & \le - 9\\ - 11x & \le - 13\\ x & \ge \frac{{13}}{{11}}\end{aligned} & \hspace{0.75in}{\rm{OR}}\hspace{0.5in} & \begin{aligned}4 - 11x & \ge 9\\ - 11x & \ge 5\\ x & \le - \frac{5}{{11}}\end{aligned}\end{align*}\]
9. \(\left| {10x + 1} \right| > - 4\)
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   This is another problem along the lines of Problems 3 and 6. However, the answer this time is VERY different. In this case we are looking for \(x\)’s that when plugged in the absolute value we will get back an answer that is greater than -4, but since absolute value only return positive numbers or zero the result will ALWAYS be greater than any negative number. So, we can plug any \(x\) we would like into this absolute value and get a number greater than -4. So, the solution to this inequality is all real numbers.